Integration Help

**Gigabyte** · February 27th 2009, 02:40 PM

How would you integrate

(2+t)^2 / 1+t^2 dt

Had a long uni break and now i'm trying to get back into it but i've forgotten a lot.

Please show intermediate steps. I've tried integration by parts, is that the right way to tackle it?

**o_O** · February 27th 2009, 03:02 PM

Note that:
$\begin{aligned}(2+t)^2 & = t^2 + 4t + 4 \\ & = {\color{red}t^2 + 1} + 4t + 3 \end{aligned}$

So: $\frac{(2+t)^2}{1+t^2} = \frac{{\color{red}t^2 + 1} + 4t + 3 }{1 + t^2} = \underbrace{\frac{{\color{red} t^2 + 1}}{1+t^2}}_{=1} + \frac{4t }{1+t^2} + \frac{3}{1+t^2}$

So your integral becomes:
$\int \frac{(2+t)^2}{1+t^2} \ dt \ \ = \ \ \int \left( 1 + \frac{4t}{1+t^2} + 3 \cdot \frac{1}{1+t^2}\right) dt$

On the RHS, use a $u$ -sub for the second integral and $\arctan (x)$ should ring a bell for the third.

**Gigabyte** · February 27th 2009, 03:56 PM

Thanks for that, the world needs more people like yourself.

Just did it and got

2 ln|1+t^2| + 3tan^-1(t) + t + C

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Thanks mate

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